Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{z^2 - 3z - 70}{z + 5} \times \dfrac{z + 5}{-9z + 90} $
Solution: First factor the quadratic. $n = \dfrac{(z - 10)(z + 7)}{z + 5} \times \dfrac{z + 5}{-9z + 90} $ Then factor out any other terms. $n = \dfrac{(z - 10)(z + 7)}{z + 5} \times \dfrac{z + 5}{-9(z - 10)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (z - 10)(z + 7) \times (z + 5) } { (z + 5) \times -9(z - 10) } $ $n = \dfrac{ (z - 10)(z + 7)(z + 5)}{ -9(z + 5)(z - 10)} $ Notice that $(z + 5)$ and $(z - 10)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(z - 10)}(z + 7)(z + 5)}{ -9(z + 5)\cancel{(z - 10)}} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $n = \dfrac{ \cancel{(z - 10)}(z + 7)\cancel{(z + 5)}}{ -9\cancel{(z + 5)}\cancel{(z - 10)}} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $n = \dfrac{z + 7}{-9} $ $n = \dfrac{-(z + 7)}{9} ; \space z \neq 10 ; \space z \neq -5 $